# Cauchy-Schwarz Problem

1. Homework Statement
Prove that ($$\sum_{j=1}^n$$ajbj)2 <= ($$\sum_{j=1}^n$$jaj2)($$\sum_{j=1}^n$$(1/j)bj2)

2. Homework Equations
Cauchy-Schwarz Inequality: |<u, v>| <= ||u||*||v||

3. The Attempt at a Solution
If I let u = $$\sum_{j=1}^n$$jaj2 and v = $$\sum_{j=1}^n$$(1/j)bj2, then I have ||u|| = sqrt(<u, u>) and ||v|| = sqrt(<v, v>). Not really sure where to go from here. Any ideas? Thanks!

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Vid
The other form is more helpful in this problem:
<u,v>^2 = <u,u><v,v>

Now, you just have to choose u and v to get the required identity. You know that <u,v> needs to equal to Sum[a_j*b_j]. You also know that the j-th term in the series on the right side is equal to the j-th component of u or v squared.

Can you finish it from here?

The other form is more helpful in this problem:
<u,v>^2 = <u,u><v,v>

Now, you just have to choose u and v to get the required identity. You know that <u,v> needs to equal to Sum[a_j*b_j]. You also know that the j-th term in the series on the right side is equal to the j-th component of u or v squared.

Can you finish it from here?
First of all, thanks for the help!

Don't you mean <u,v>2 <= <u,u><v,v> ? Also, I'm not sure what you mean by your last sentence. "The series on the right side" is referring to which series? The product of the two series? Anyway, how do I know that? Sorry for these possibly stupid questions, and thanks again for your help.

Vid
Yea, that should be less than or equal to.

I meant that one series was for u and the other for v. With the standard dot product <u,u> = Sum[(u_j)^2].

So Sum[u_j^2] = Sum[j*a_j^2]
and Sum[v_j^2] = Sum[(1/j)*b_j^2]

What does this tell us about what u and v must be equal to?

Yea, that should be less than or equal to.

I meant that one series was for u and the other for v. With the standard dot product <u,u> = Sum[(u_j)^2].

So Sum[u_j^2] = Sum[j*a_j^2]
and Sum[v_j^2] = Sum[(1/j)*b_j^2]

What does this tell us about what u and v must be equal to?
I don't think I'm seeing it. u_j = sqrt(j)*a_j and u = Sum[u_j] = Sum[sqrt(j)*a_j] ?
v_j = (1/sqrt(j))*b_j and v = Sum[v_j] = Sum[(1/sqrt(j))*b_j] ?

Is this correct? I don't really see how this helps.

the cs inequality says <p, q> <= |p||q|, here | | denotes the norm

so (<p, q>)^2 <= |p|^2|q|^2

Set p = (a_1/sqrt(1),..., a_n/sqrt(n)), q= (sqrt(1)*b_1, 2*b_2,..., sqrt(n)*b_n)

so your sum on the left = (<p, q>)^2 <= |p|^2|q|^2 = the stuff you have on the right

just verify this

Ok, just to make sure I got it:

Cauchy Schwarz says |<u, v>| <= ||u||*||v||. This implies <u, v>2 <= <u, u><u, v>. Let <u, u> = $$\sum$$uj2 = $$\sum$$jaj2 and let <v, v> = $$\sum$$vj2 = $$\sum$$(1/j)bj2.

Then we have u = $$\sum$$uj = $$\sum$$sqrt(j)*aj and v = $$\sum$$vj = $$\sum$$(1/sqrt(j))*bj.

Then we have <u, v> = $$\sum$$ujvj = ... = $$\sum$$ajbj. Then square both sides to get <u, v>2 = ($$\sum$$ajbj)2.

Then after plugging in, this completes the proof. Thanks again for all your help guys!